Show that if is open, then can be written as an at most countable union of disjoint intervals, i.e., . (It's possible that or for some .) Hint: One way to do this is to put open intervals around each rational point in E in such a way that every point of E and only points of E are contained somewhere in these intervals. Then combine the intervals that intersect.
OK, intuitively, I get this, but what confuses me about the method they suggest is when I'm told to combine the intervals that intersect. Doesn't that imply that they aren't disjoint?! At firstI had the idea to take the set of all Neigborhoods of all the rationals in E of rational radius that do not intersect the complement of E, but these are not disjoint. Then I decided to take an arbitrarily large neighborhood in E and then take neighborhoods of the space that is left over and keep filling in the gaps with more and more neighborhoods until I have an at most countable amount of neighborhoods that are dense in E, but I had trouble getting that down and I also figured I should probably utilize the hint... But the hint confuses me more than the problem statement.